How do you differentiate #log_(3)x #?

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Answer 1 · 2017-01-04T08:58:02

#(dy)/(dx)=1/(xln3)#

We need to change it to natural logs before differentiating

#y=log_3x=>3^y=x#

#ln3^y=lnx#

#yln3=lnx#

differentiate wrt #x#

#(dy)/(dx)ln3=1/x#

#(dy)/(dx)=1/(xln3)#